Q.1 Solve the following pair of linear equations by the elimination method and the substitution method :
- \(x + y = 5 \text{ and } 2x – 3y = 4\)
- \(3x + 4y = 10 \text{ and } 2x – 2y = 2\)
- \(3x – 5y – 4 = 0 \text{ and } 9x = 2y + 7\)
- \(\frac{x}{2} + \frac{2y}{3}\text{ and } x-\frac{y}{3}=3\)
- \(x + y = 5 \text{ and } 2x – 3y = 4\)
Solution:
Given the pair of linear equations \(x + y = 5\) and \(2x - 3y = 4\), the solution is found first by the elimination method and then by the substitution method to check the result.
By elimination method:
The given equations are
$$x + y = 5\tag{1}$$ $$2x - 3y = 4\tag{2}$$Multiply equation (1) by \(2\) to make the coefficients of \(x\) equal.
$$ \begin{align} (x + y)\times 2 &\\\Rightarrow 2x + 2y = 10\tag{3} \end{align} $$Subtract equation (3) from equation (2).
$$ \begin{aligned} 2x - 3y - (2x + 2y) &= 4 - 10\\ 2x - 3y - 2x - 2y &= -6\\ -5y &= -6\\ y &= \dfrac{6}{5} \end{aligned} $$Substitute \(y = \dfrac{6}{5}\) in equation (1).
$$ \begin{aligned} x + y &= 5\\ x + \dfrac{6}{5} &= 5 \end{aligned} $$Multiply both sides by \(5\) to remove the denominator.
$$ \begin{aligned} 5x + 6 &= 25\\ 5x &= 25 - 6\\ 5x &= 19\\ x &= \dfrac{19}{5} \end{aligned} $$Hence, by the elimination method, the solution is \(x = \dfrac{19}{5}\) and \(y = \dfrac{6}{5}\).
Solving by substitution method
Start again with the first equation and express \(x\) in terms of \(y\).
$$ \begin{align} x + y &= 5\\ x &= 5 - y\tag{1} \end{align} $$ $$2x - 3y = 4\tag{2}$$Substitute \(x = 5 - y\) in equation (2).
$$ \begin{aligned} 2(5 - y) - 3y &= 4\\ 10 - 2y - 3y &= 4\\ 10 - 5y &= 4\\ -5y &= 4 - 10\\ -5y &= -6\\ y &= \dfrac{6}{5} \end{aligned} $$Now put \(y = \dfrac{6}{5}\) back in \(x + y = 5\).
$$ \begin{aligned} x + y &= 5\\ x + \dfrac{6}{5} &= 5 \end{aligned} $$Multiply both sides by \(5\).
$$ \begin{aligned} 5x + 6 &= 25\\ 5x &= 25 - 6\\ 5x &= 19\\ x &= \dfrac{19}{5} \end{aligned} $$Thus, by the substitution method also, the solution of the pair of linear equations is \(x = \dfrac{19}{5}\) and \(y = \dfrac{6}{5}\), confirming the result obtained by the elimination method. [2]
- \(3x + 4y = 10 \text{ and } 2x – 2y = 2\)
Solution:
By elimination method:
$$3x+4y=10\tag{1}$$ $$2x-2y=2y\tag{2}$$Multiply eqn-2 by 2
$$\begin{align}\left( 2x-2y=2\right) \times 2\\ 4x-4y=4\tag{3}\end{align}$$Adding Equation-1 and equation-3
$$\begin{aligned}\left( 3x+4y\right) +\left( 4x-4y\right) &=10+4\\ 3x+4y+4x-4y&=14\\ 7x&=14\\ x&=\dfrac{14}{7}\\ &=2\end{aligned}$$Substituting \(x = 2\) in equation-1
$$\begin{aligned}3x+4y&=10\\ 3\times 2+4y&=10\\ 6+4y&=10\\ 4y&=10-6\\ 4y&=4\\ y&=\dfrac{4}{4}\\ y&=1\end{aligned}$$Solution by substitution Method.
$$3x+4y=10$$ $$3x=10-4y$$ $$x=\dfrac{1}{3}\left[ 10-4y\right]\tag{1} $$ $$2x-2y=2\tag{2}$$substituting value of \(x\) in eqn-(2)
$$\begin{aligned}2x-2y&=2\\ 2\times \dfrac{1}{3}\left[ 10-4y\right] -2y&=2\end{aligned}$$Multiplying both side by 3
$$\begin{aligned}2\left( 10-4y\right) -6y&=6\\ 20-8y-6y&=6\\ -14y&=6-20\\ &=14\\ y&=\dfrac{14}{14}\\ &=1\end{aligned}$$Putting \(y = 1\) in eqn- (1)
$$\begin{aligned}3x+4y&=10\\ 3x+4&=10\\ 3x&=10-4\\ x&=\dfrac{6}{3}\\ &=2\end{aligned}$$
- \(3x – 5y – 4 = 0 \text{ and } 9x = 2y + 7\)
Solution:
By elimination method:
$$3x-5y-4=0\tag{1}$$ $$\begin{align}9x&=2y+7\\ \Rightarrow 9x-2y-7&=0\tag{2}\end{align}$$Multiply equation-(1) by 3
$$\begin{align} (3x-5y-4&=0) \times 3\\ 9x-15y-12&=0\tag{3}\end{align}$$Subtracting eqn-2 by equation-3
$$\scriptsize\begin{aligned}9x-2y-7-\left( 9x-15y-12\right) &=0\\ 9x-2y-7-9x+15y+12&=0\\ 13y+5&=0\\ 13y&=-5\\ y&=\dfrac{-5}{13}\end{aligned}$$Substituting value of \(y\) in equation-(1)
$$\begin{aligned}3x-5y-4&=0\\ 3x-5\left( \dfrac{-5}{13}\right) -4&=0\end{aligned}$$Multiplying both side by 13
$$\begin{aligned} ( 3x+\dfrac{25}{13}-4&=0) \times 13\\ 39x+25-52&=0\\ 39x-27&=0\\ 39x&=27\\ x&=\dfrac{27}{39}\\ x&=\dfrac{9}{13}\\\\ \boxed{x=\dfrac{9}{13},y=-\dfrac{5}{13}}\end{aligned}$$Solving by substitution Method
$$\begin{align}3x-5y-4&=0\\ 3x&=5y+4\\ x&=\dfrac{1}{3}\left[ 5y+4\right] \tag{1}\end{align}$$ $$9x=2y+7\tag{2}$$Substituting value of \(x\) in equation-(2)
$$\begin{aligned}9x&=2y+7\\ 9\times \dfrac{1}{3}\left[ 5y+4\right] &=2y+7\\ 3\left( 5y+4\right) &=2y+7\\ 15y+12-2y&=7\\ 13y&=7-12\\ y&=-\dfrac{5}{13}\end{aligned}$$Putting value of \(y\) in equation-(1)
$$\begin{aligned}3x-3y-4&=0\\ 3x-5\left( \dfrac{-5}{13}\right) -4&=0\end{aligned}$$Multiplying both side by 13
$$\begin{aligned}39x+25-52&=0\\ 39x-27&=0\\ 39x&=27\\ x&=\dfrac{27}{39}\\ x&=\dfrac{9}{13}\\\\ \boxed{x=\dfrac{9}{13},\ y=-\dfrac{5}{13}}\end{aligned}$$
- \(\frac{x}{2} + \frac{2y}{3}\text{ and } x-\frac{y}{3}=3\)
Solution:
By elimination method:
$$\dfrac{x}{2}+\dfrac{2y}{3}=-1$$Simplifying
$$3x+4y=-6\tag{1}$$ $$x-\dfrac{y}{3}=3$$Multiplying both side by 3
$$3x-y=9\tag{2}$$Subtracting equation-(1) by equation-(2)
$$\begin{aligned}3x-y-\left( 3x+4y\right) &=9-\left( -6\right) \\ 3x-y-3x-4y&=15\\ -5y&=15\\ y&=\dfrac{15}{-5}\\ y&=-3\end{aligned}$$putting \(y=-3\) in equation-(1)
$$\scriptsize\begin{aligned}3x+4y&=-6\\ 3x+4\left( -3\right) &=-6\\ 3x-12&=-6\\ 3x&=12-6\\ 3x&=6\\ x&=\dfrac{6}{3}\\ &=2\\\\ \boxed{x=2,\ y=-3}\end{aligned}$$Solving by substitution Method
$$\dfrac{x}{2}+\dfrac{2y}{3}=-1$$Simplifying
$$\begin{align}3x+4y=-6\\ 3x=-6-4y\\ x=\dfrac{1}{3}\left( -6-4y\right) \tag{1}\end{align}$$ $$x-\dfrac{y}{3}=3$$Multiplying both side by 3
$$3x-y=9\tag{2}$$Substituting value of \(x\) from equation-(1) to equation-(2)
$$\begin{aligned}3x-y&=9\\ 3\times \dfrac{1}{3}\left( -6-4y\right) -y&=9\\ -6-4y-y&=9\\ -5y&=9+6\\ -5y&=15\\ y&=-\dfrac{15}{5}\\ &=-3\end{aligned}$$Putting \(y =3\) in equation-(1)
$$\begin{aligned}x&=\dfrac{1}{3}\left[ -6-4y\right] \\ &=\dfrac{1}{3}\left[ -6-4\times \left( -3\right) \right] \\ &=\dfrac{1}{3}\left[ -6+12\right] \\ &=\dfrac{1}{3}\times 6\\ &=2\\\\ \boxed{x=2,\ y=-3}\end{aligned}$$
Q.2 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :
-
If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces
to 1. It becomes \(\frac{1}{2}\) if we only add 1 to the denominator. What is the fraction?
Solution:
Let fraction be $$\dfrac{x}{y}$$Adding 1 to numerator and subtracting 1 to denominator fraction reduces to 1,
$$\begin{align}\dfrac{x+1}{y-1}&=1\\ x+1&=y-1\\ x-y&=-2\tag{1}\end{align}$$
thusFraction become \(\frac{1}{2}\) if 1 added to denominator,
$$\begin{align}\dfrac{x}{y+1}&=\dfrac{1}{2}\\ 2x&=y+1\\ 2x-y&=1\tag{2}\end{align}$$
thusSubtracting equation-(2) from equation-(1)
$$\begin{aligned}2x-y-\left( x-y\right) &=1-\left( -2\right) \\ 2x-y-x+y&=3\\ x&=3\end{aligned}$$Substituting \(x = 3\) in equation-(1)
$$\begin{aligned}x-y&=-2\\ 3-y&=-2\\ -y&=-2-3\\ -y&=-5\\ y&=5\\\\ \boxed{x=3,\ y=5}\end{aligned}$$ Hence, Fraction is: $$\dfrac{x}{y}=\dfrac{3}{5}$$
-
Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as
old as Sonu. How old are Nuri and Sonu?
Solution:
Let Nuri present age = \(x\)
$$\begin{align} x-5&=3\left( y-5\right) \\ x-5&=3y-15\\ x-3y&=-15+5\\ x-3y&=-10\tag{1}\end{align}$$
and Sonu present age = \(y\)
Five year ago Nuri age = \(x-5\)
Sonu Age = \(y-5\)
Five years ago Nuri was thirce as old as Sonu.
thus,10 year hence,
$$\begin{align} \left( x+10\right) &=2\left( y+10\right) \\ x+10&=2y+20\\ x-2y&=20-10\\ x-2y&=10\tag{2}\end{align}$$
Nuri age = \(x+10\)
Sonu age = \(y+10\)
after 10 year Nuri will be twice as old as Sonu
thus,Subtracting equation-(2) from equation-(1)
$$\begin{aligned} x-3y-\left( x-2y\right) &=-10-\left( 10\right) \\ x-3y-x+2y&=-20\\ -y&=-20\\ y&=20\end{aligned}$$Substituting value of \(y\) in equation-(1)
$$\begin{aligned} x-3y&=-10\\ x-3\left( 20\right) &=-10\\ x-60&=-10\\ x&=60-10\\ &=50 \end{aligned}$$Present age of Nuri is 50 years and Sonu is 20 years.
-
The sum of the digits of a two-digit number is 9. Also, nine times this number is
twice the number obtained by reversing the order of the digits. Find the number.
Solution:
Let first digit of two digit number is \(x\)
$$x+y=9\tag{1}
and the second digit = \(y\)
Given that the sum of digits is 9,
thuslet two digit number be
$$10x+y$$Number when digits will be reversed
$$10y+x$$nine times \(10x+y\) is twice the \(10y+x\) obtained by reversing the order of the digits,
$$\begin{align} 9\left( 10x+y\right) &=2\left( 10y+x\right) \\ 90x+9y&=20y+2x\\ 88x&=11y\\ 8x&=y\\ y&=8x\tag{2}\end{align}$$
thusSubstituting value of \(y\) in equation
$$\begin{align} x+y&=9\\ x+8x&=9\\ 9x&=9\\ x&=1\tag{2}\end{align}$$Substituting (\x=1\) in equation-(2)
$$\begin{aligned}y=8x\\ y=8\times 1\\ =8\\\\ \boxed{x=1,\ y=8}\end{aligned}$$Hence Number
$$\begin{aligned}10x+y\\ 10+8\\ =18\end{aligned}$$
-
Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her
₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of
₹ 50 and ₹ 100 she received.
Solution:
Let ₹50 notes be \(x\) and ₹100 notes be \(y\),
$$50x+100y=2000\tag{1}$$
then value sum of notes is:Total count of Notes:
$$\begin{align}x+y&=25\\ x&=25-y\tag{2}\end{align}$$Substituting value of \(x\) in equation-(1)
$$\begin{aligned}50x+100y&=2000\\ 50\left( 25-y\right) +100y&=2000\\ 1250-50y+100y&=2000\\ 50y&=2000-1250\\ &=750\\ 50y&=750\\ y&=\dfrac{75}{50}\\ y&=15\end{aligned}$$Substituting value of \(y\) in equation-2
$$\begin{aligned}x+y&=25\\ x+15&=25\\ x&=25-15\\ x&=10\end{aligned}$$Hence, ₹50 notes = 10 and ₹100 notes = 15
-
A lending library has a fixed charge for the first three days and an additional charge
for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy
paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge
for each extra day.
Solution:
Let fixed charges of library for three days be \(x\)
$$x+4y=27\tag{1}$$
additional charges at the rate of \(y/\text{day}\)
therefore, Rent for 7 daysRent for 5 days
$$x+2y=21\tag{2}$$Subtracting equation-(2) from equation-(1)
$$\begin{aligned}x+4y-\left( x+2y\right) &=27-21\\ x+4y-x-2y&=6\\ 2y&=6\\ y&=\dfrac{6}{2}\\ &=3\end{aligned}$$Substituting value of \(y\) in equation-(1)
$$\begin{aligned} x+4y&=27\\ x+4\times 3&=27\\ x+12&=27\\ x&=27-12\\ x&=15\end{aligned}$$ $$\boxed{x=15,\ y=3}$$Fixed charges = ₹15 and charges for extra days is ₹3/day